Chapter 1: Vector Spaces
Linear Algebra Done Right, by Sheldon Axler
A:
R and C
Problem 1
Suppose a and b are real numbers, not both 0. Find real numbers c and
d such that
1
= c + di
a + bi
Proof. We have
1
a − bi
a
b
= 2
= 2
− 2
i,
a + bi
a + b2
a + b2
a + b2
and hence let
c=
a
,
a2 + b2
d=−
b
,
a2 + b2
and we’re done.
Problem 3
Find two distinct square roots of i.
Proof. Suppose a, b ∈ R are such that (a + bi)2 = i. Then
(a2 − b2 ) + (2ab)i = i.
Since the real and imaginary part of both sides must be equal, respectively, we
have a system of two equations in two variables
a2 − b2 = 0
1
ab = .
2
The first equation implies b = ±a. Plugging b = −a into the second equation
would imply −a2 = 1/2, which is
√ impossible, and hence we must have a = b.
But this in turn tells us a = ±1/ 2, and hence our two roots are
1
± √
(1 + i),
2
as desired.
1
Problem 5
Show that (α + β) + λ = α + (β + λ) for all α, β, λ ∈ C.
Proof. Suppose α = a1 + a2 i, β = b1 + b2 i, and λ = c1 + c2 i for ak , bk , ck ∈ R,
where k = 1, 2. Then
(α + β) + λ = (a1 + a2 i) + (b1 + b2 i) + (c1 + c2 i)
= (a1 + b1 ) + (a2 + b2 )i + (c1 + c2 i)
= (a1 + b1 ) + c1 + (a2 + b2 ) + c2 i
= a1 + (b1 + c1 ) + a2 + (b2 + c2 ) i
= (a1 + a2 i) + (b1 + c1 ) + (b2 + c2 )i
= (a1 + a2 i) + (b1 + b2 i) + (c1 + c2 )i
= α + (β + λ),
as desired.
Problem 7
Show that for every α ∈ C, there exists a unique β ∈ C such that
α + β = 0.
Proof. Suppose α = a1 + a2 i for some a1 , a2 ∈ R, and define β = −a1 − a2 i.
Then
α + β = (a1 + a2 i) + (−a1 − a2 i)
= (a1 − a1 ) + (a2 − a2 )i
= 0 + 0i
= 0,
proving existence. To see that β is unique, suppose λ ∈ C such that α + λ = 0.
Then
λ = λ + (α + β) = (λ + α) + β = 0 + β = β,
and thus β is unique.
Problem 9
Show that λ(α + β) = λα + λβ for all λ, α, β ∈ C.
2
Proof. Suppose α = a1 + a2 i, β = b1 + b2 i, and λ = c1 + c2 i for ak , bk , ck ∈ R,
where k = 1, 2. Then
λ(α + β) = (c1 + c2 i) (a1 + a2 i) + (b1 + b2 i)
= (c1 + c2 i) (a1 + b1 ) + (a2 + b2 )i
= c1 (a1 + b1 ) − c2 (a2 + b2 ) + c1 (a2 + b2 ) + c2 (a1 + b1 ) i
= (c1 a1 + c1 b1 ) − (c2 a2 + c2 b2 ) + (c1 a2 + c1 b2 ) + (c2 a1 + c2 b1 ) i
= (c1 a1 − c2 a2 ) + (c1 b1 − c2 b2 ) + (c1 a2 + c2 a1 ) + (c1 b2 + c2 b1 ) i
= (c1 a1 − c2 a2 ) + (c1 a2 + c2 a1 )i + (c1 b1 − c2 b2 ) + (c1 b2 + c2 b1 )i
= (c1 + c2 i)(a1 + a2 i) + (c1 + c2 i)(b1 + b2 i)
= λα + λβ,
as desired.
Problem 11
Explain why there does not exist λ ∈ C such that
λ(2 − 3i, 5 + 4i, −6 + 7i) = (12 − 5i, 7 + 22i, −32 − 9i).
Proof. Suppose such a λ ∈ C exists, say λ = a + bi for some a, b ∈ R. Then
(a + bi)(2 − 3i) = 12 − 5i
(a + bi)(5 + 4i) = 7 + 22i
(a + bi)(−6 + 7i) = −32 − 9i,
which is equivalent to
(2a + 3b) + (−3a + 2b)i = 12 − 5i
(1)
(5a − 4b) + (4a + 5b)i = 7 + 22i
(2)
(−6a − 7b) + (7a − 6b)i = −32 − 9i.
(3)
For each equation above, the real part of the LHS must equal the real part of
the RHS, and similarly for their imaginary parts. In particular, the following
two equations hold by comparing the real parts of Equations (1) and (3)
2a + 3b = 12
−6a − 7b = −32.
Multiplying the first equation by 3 and adding it to the second, we find b = 2.
Substituting this value back into the first equation yields a = 2. However,
comparing the imaginary parts of Equation (3) tells us we must have
7a − 6b = −9,
a contradiction, since a = 3 and b = 2 yields 7a − 6b = 9. Thus no such λ ∈ C
exists, as was to be shown.
3
Problem 13
Show that (ab)x = a(bx) for all x ∈ Fn and all a, b ∈ F.
Proof. We may write x = (x1 , . . . , xn ) for x1 , . . . , xn ∈ F. It follows
(ab)x = (ab)x1 , . . . , (ab)xn
= a(bx1 ), . . . , a(bxn )
= a (bx1 , . . . , bxn )
= a(bx),
as desired.
Problem 15
Show that λ(x + y) = λx + λy for all λ ∈ F and all x, y ∈ Fn .
Proof. We may write x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) for xk , yk ∈ Fn ,
where k = 1, . . . , n. It follows
λ(x + y) = λ((x1 , . . . , xn ) + (y1 , . . . , yn ))
= λ((x1 + y1 ) + · · · + (xn + yn ))
= (λ(x1 + y1 ) + · · · + λ(xn + yn ))
= ((λx1 + λy1 ) + · · · + (λxn + λyn ))
= (λx1 + · · · + λxn ) + (λy1 + · · · + λyn )
= λx + λy,
as desired.
B:
Definition of a Vector Space
Problem 1
Prove that −(−v) = v for every v ∈ V .
Proof. We wish to show that v is the additive inverse of (−v). We have
(−v) + v = (−1)v + 1v = (−1 + 1)v = 0v = 0,
as desired.
Problem 3
Suppose v, w ∈ V . Explain why there exists a unique x ∈ V such that
v + 3x = w.
4
Proof. First we prove existence. Define x ∈ V by
x=
1
(w − v).
3
Then
1
(w − v)
v + 3x = v + 3
3
1
=v+ 3·
(w − v)
3
= v + (w − v)
= w,
and so such an x exists. To see that it’s unique, suppose y ∈ V such that
v + 3y = w. Then
v + 3y = v + 3x ⇐⇒ 3y = 3x ⇐⇒ y = x,
proving uniqueness.
Problem 5
Show that in the definition of a vector space (1.19), the additive inverse
condition can be replaced with the condition that
0v = 0 for all v ∈ V.
Here the 0 on the left side is the number 0, and the 0 on the right side is
the additive identity of V . (The phrase “a condition can be replaced” in
a definition means that the collection of objection satisfying the definition
is unchanged if the original condition is replaced with the new definition.)
Proof. We show that the two statements are equivalent.
First suppose that every v ∈ V has an additive inverse. Since we have
0v + 0v = (0 + 0)v = 0v,
adding the additive inverse to both sides yields 0v = 0.
Conversely, suppose that 0v = 0 for all v ∈ V . Then
v + (−1)v = (1 + (−1))v = 0v = 0,
and hence every element has an additive inverse, as desired.
5
C:
Subspaces
Problem 1
For each of the following subsets of F3 , determine whether it is a subspace
of F3 :
1. {(x1 , x2 , x3 ) ∈ F3 | x1 + 2x2 + 3x3 = 0}
2. {(x1 , x2 , x3 ) ∈ F3 | x1 + 2x2 + 3x3 = 4}
3. {(x1 , x2 , x3 ) ∈ F3 | x1 x2 x3 = 0}
4. {(x1 , x2 , x3 ) ∈ F3 | x1 = 5x3 }
Proof. (a) Let S denote the specified subset. We claim S is a subspace. To
see this, note that 0 + 2 · 0 + 3 · 0 = 0, and hence 0 ∈ S. Now suppose
x = (x1 , x2 , x3 ) ∈ S and y = (y1 , y2 , y3 ) ∈ S. Then
x1 + 2x2 + 3x3 = 0
and y1 + 2y2 + 3y3 = 0,
and hence
(x1 + 2x2 + 3x3 ) + (y1 + 2y2 + 3y3 ) = (x1 + y1 ) + 2(x2 + y2 ) + 3(x3 + y3 )
= 0,
and so x + y ∈ S and S is closed under addition. Now letting a ∈ F, we
have
a(x1 + 2x2 + 3x3 ) = ax1 + 2(ax2 ) + 3(ax3 ) = 0,
and hence ax ∈ S as well, and so S is closed under scalar multiplication,
thus proving S is a subspace, as claimed.
(b) Let S denote the specified subset. Then S is not a subspace, for 0 + 2 · 0 +
3 · 0 = 0, and hence S does not contain the additive identity.
(c) Let S denote the specified subset. We claim S is not a subspace since it is
not closed under addition. To see this, let x = (1, 0, 0) and y = (0, 1, 1).
Then x, y ∈ S, but x + y = (1, 1, 1) 6∈ S since (x1 + y1 )(x2 + y2 )(x3 + y3 ) =
1 · 1 · 1 6= 0.
(d) Let S denote the specified subset. We claim S is a subspace. To see this,
note that 0 = 5 · 0, and hence 0 ∈ S. Now suppose x = (x1 , x2 , x3 ) ∈ S
and y = (y1 , y2 , y3 ) ∈ S. Then
x1 = 5x3
and
y1 = 5y3 ,
and hence
(x1 + y1 ) = 5x3 + 5y3 = 5(x3 + y3 ),
6
and so x + y ∈ S and S is closed under addition. Now letting a ∈ F, we
have
a(x1 ) = a(5x3 )
and thus
(ax1 ) = 5(ax3 ),
showing ax ∈ S as well. Therefore S is closed under scalar multiplication
as well, proving S is a subspace, as claimed.
Problem 3
Show that the set of differentiable real-valued functions f on the interval
(−4, 4) such that f 0 (−1) = 3f (2) is a subspace of R(−4,4) .
Proof. Let S denote the set of differentiable real-valued functions f on the
interval (−4, 4) such that f 0 (−1) = 3f (2) . Denote the zero-function (the
additive identity of R(−4,4) ) by f0 . Then f00 = f0 and f00 (−1) = 0, and hence
f00 (−1) = 3f0 (2) — both sides are 0 — showing f0 ∈ S. Now suppose f, g ∈ S.
Then
(f + g)0 = f 0 + g 0 ,
and hence
(f + g)0 (−1) = f 0 (−1) + g 0 (−1)
= 3f (2) + 3g(2)
= 3(f (2) + g(2))
= 3(f + g)(2),
showing (f + g) ∈ S and S is closed under addition. Now letting a ∈ R, we have
a(f 0 (−1)) = a(3f (2)) =⇒ (af 0 )(−1) = 3(af )(2),
and hence (af ) ∈ S as well, and S is closed under scalar multiplication. Therefore,
S is a subspace.
Problem 5
Is R2 a subspace of the complex vector space C2 ?
Proof. The set R2 is not a subspace of C2 over the field C since R2 is not
closed under scalar multiplication. In particular, we have ix 6∈ R2 for all
x ∈ R2 − {0}.
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Problem 7
Give an example of a nonempty subset U of R2 such that U is closed
under addition and taking additive inverses (meaning −u ∈ U whenever
u ∈ U ), but U is not a subspace of R2 .
Proof. Consider the set Z × Z. Then for all (a, b) ∈ Z × Z, we have (−a, −b) ∈
Z × Z, and so it’s closed under additive inverses. Similarly, for any (c, d) ∈ Z × Z,
we have (a, b) + (c, d) = (a + c, b + d) ∈ Z × Z, and so it’s closed under addition.
But Z×Z is not a subspace of R2 , since it is not closed under scalar multiplication.
In particular, 21 (1, 1) = 12 , 12 6∈ Z × Z.
Problem 9
A function f : R → R is called periodic if there exists a positive number
p such that f (x) = f (x + p) for all x ∈ R. Is the set of periodic functions
from R to R a subspace of RR ? Explain.
Proof. Let P denote the set of periodic functions from R to R. We claim P is
not a subspace of RR , since it is not closed under addition. To see this, define
2π
f (x) = cos √ x
and g(x) = cos(2πx).
2
Then
f (x +
and so f has period
√
√
√
2π
√ (x + 2)
2
2π
= cos √ x + 2π
2
2π
= cos √ x
2
= f (x),
2) = cos
2, and
g(x + 1) = cos(2π(x + 1)) = cos(2πx + 2π) = cos(2πx) = g(x),
so that g has period 1.
Suppose by way of contradiction that f + g were periodic with respect to
some p ∈ R+ . Then, since
2π
(f + g)(0) = cos √ · 0 + cos(2π · 0) = cos(0) + cos(0) = 2,
2
8
by periodicity of f + g we must also have
2π
(f + g)(p) = cos √ p + cos(2πp) = 2.
2
The maximum of cosine is 1, and hence both f and g must have maxima at p.
But the maxima of cosine occur at the integer multiples of 2π, and hence we
must have
2π
√ p = 2πn and
2
2πp = 2πm
for some n, m ∈ Z+ . But this implies
√
p = 2n and p = m.
In other words,
m √
= 2,
n
√
a contradiction since 2 is irrational. Thus f + g cannot be periodic, and indeed
P is not closed under addition, as claimed.
Problem 11
Prove that the intersection of every collection of subspaces of V is a
subspace of V .
Proof. Let C denote a collection of subspaces of V , and let
\
U=
W.
W ∈C
Then, since 0 ∈ W for all W ∈ C, we have 0 ∈ U and so U contains the additive
identity. Now suppose u, v ∈ U . Then u, v ∈ W for all W ∈ C, and hence
u + v ∈ W for all W ∈ C. Therefore, u + v ∈ U and U is closed under addition.
Next let a ∈ F. Then au ∈ W for all W ∈ C, and hence au ∈ U , showing U is
closed under scalar multiplication. Therefore, U is indeed a subspace of V .
Problem 12
Prove that the union of two subspaces of V is a subspace of V if and only
if one of the subspaces is contained in the other.
Proof. Let U1 , U2 be subspaces of V .
First suppose that one of the subspaces is contained in the other. Then either
U1 ∪ U2 = U1 or U1 ∪ U2 = U2 , and in both cases U1 ∪ U2 is indeed a subspace
of V .
9
Conversely, suppose by way of contradiction that U1 ∪ U2 is a subspace of
V , but neither subspace is contained in the other. That is, the sets U1 \ U2 and
U2 \ U1 are both nonempty. Let x ∈ U1 \ U2 and y ∈ U2 \ U1 . We claim x + y 6∈ U1
and x + y 6∈ U2 , so that U1 ∪ U2 is not closed under addition, a contradiction. To
see this, suppose x + y ∈ U1 . Then (x + y) − x ∈ U1 by closure of addition in U1 ,
but this is absurd since this implies y ∈ U1 , contrary to assumption. Similarly,
suppose x + y ∈ U2 . Then (x + y) − y ∈ U2 , which is also absurd since this
implies x ∈ U2 . Therefore U1 ∪ U2 is not closed under addition, producing a
contradiction as claimed. Thus we must have one of the subspaces contained in
the other, as desired.
Problem 13
Prove that the union of three subspaces of V is a subspace of V if and
only if one of the subspaces contains the other two.
Proof. Let U1 , U2 , U3 be subspaces of V .
(⇐) Suppose that one of the subspaces contains the other two. Without loss
of generality, assume U1 ⊆ U3 and U2 ⊆ U3 . Then U1 ∪ U2 ∪ U3 = U3 , and so
U1 ∪ U2 ∪ U3 is indeed a subspace of V .
(⇒) Now suppose U1 ∪U2 ∪U3 is a subspace. If U2 contains U3 (or conversely),
let W = U2 ∪ U3 . Then applying Problem 12 to the union U1 ∪ W , we have that
either U1 contains W or W contains U1 , showing that one of the three subspaces
contains the other two, as desired. So assume U2 and U3 are such that neither
contains the other. Let
x ∈ U2 \ U3
and y ∈ U3 \ U2 ,
and choose a, b ∈ F \ {0} such that a − b = 1 (such a, b exist since we assume F
is not finite).
We claim that ax + y and bx + y are both in U1 . To see that ax + y ∈ U1 ,
suppose not. Then either ax + y ∈ U2 or ax + y ∈ U3 . If ax + y ∈ U2 , then
we have (ax + y) − ax = y ∈ U2 , a contradiction. And if ax + y ∈ U3 , we have
(ax + y) − y = ax ∈ U3 , another contradiction, and so ax + y ∈ U1 . Similarly
for bx + y, suppose bx + y ∈ U2 . Then (bx + y) − bx = y ∈ U2 , a contradiction.
And if bx + y ∈ U3 , then (bx + y) − y = bx ∈ U3 , also a contradiction. Thus
bx + y ∈ U1 as well. Therefore
(ax + y) − (bx + y) = (a − b)x = x ∈ U1 .
Now, since x ∈ U2 \U3 implies x ∈ U1 , we have U2 \U3 ⊆ U1 . A similar argument
shows that x + ay and x + by must be in U1 as well, and hence
(x + ay) − (x + by) = (a − b)y = y ∈ U1 ,
and therefore U3 \ U2 ⊆ U1 . If U2 ∩ U3 = ∅, we’re done, so assume otherwise.
10
Now for any u ∈ U2 ∩ U3 , choose v ∈ U3 \ U2 ⊆ U1 . Then u + v 6∈ U2 ∩ U3 ,
for otherwise (u + v) − u = v ∈ U2 , a contradiction. But this implies u + v must
be in U1 , and hence so is (u + v) − v = u. In other words, if u ∈ U2 ∩ U3 , then
u ∈ U1 , and hence U2 ∩ U3 ⊆ U1 , as was to be shown.
Problem 15
Suppose U is a subspace of V . What is U + V ?
Proof. We claim U + V = V . First suppose x ∈ V . Then x = 0 + x ∈ U + V ,
and hence V ⊆ U + V . Now suppose y ∈ U + V . Then there exist u ∈ U and
v ∈ V such that y = u + v. But since U is a subspace of V , we have u ∈ V , and
hence u + v ∈ V . Therefore U + V ⊆ V , proving the claim.
Problem 17
Is the operation of addition on the subspaces of V associative? In other
words, if U1 , U2 , U3 are subspaces of V , is
(U1 + U2 ) + U3 = U1 + (U2 + U3 )?
Proof. Let U1 , U2 , U3 be subspaces of V , and let V1 = U1 + U2 and V2 = U2 + U3 .
We claim
V1 + U3 = U1 + V2 .
To see this, suppose x ∈ V1 + U3 . Then there exist v1 ∈ V1 and u3 ∈ U3 such
that x = v1 + u3 . But since v1 ∈ V1 = U1 + U2 , there exist u1 ∈ U1 and u2 ∈ U2
such that v1 = u1 + u2 . Then x = u1 + u2 + u3 , and since u2 + u3 ∈ U2 + U3 = V2 ,
we have x ∈ U1 + V2 and hence V1 + U3 ⊆ U1 + V2 . Now suppose y ∈ U1 + V2 .
Then there exist u01 ∈ U1 and v2 ∈ V2 such that y = u01 + v2 . But since
v2 ∈ V2 = U2 + U3 , there exist u02 ∈ U2 and u03 ∈ U3 such that v2 = u02 + u03 .
Then y = u01 + u02 + u03 , and since u01 + u02 ∈ U1 + U2 = V1 , have y ∈ V1 + U3 and
hence U1 + V2 ⊆ V1 + U3 . Thus V1 + U3 = U1 + V2 , as claimed.
Problem 19
Prove or give a counterexample: if U1 , U2 , W are subspaces of V such
that
U1 + W = U2 + W,
then U1 = U2 .
Proof. The statement is false. To see this, let V = U1 = W = R2 and U2 =
R × {0}. Then U1 + W = R2 and U2 + W = R2 , but clearly U1 6= U2 .
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Problem 21
Suppose
U = {(x, y, x + y, x − y, 2x) ∈ F5 | x, y ∈ F}.
Find a subspace W of F5 such that F5 = U ⊕ W .
Proof. Let v1 = (1, 0, 1, 1, 2, ), v2 = (0, 1, 1, −1, 0), so that we may instead write
V as
V = {α1 v1 + α2 v2 ∈ F5 | α1 , α2 ∈ F}.
Now let w1 = (0, 0, 1, 0, 0), w2 = (0, 0, 0, 1, 0), w3 = (0, 0, 0, 0, 1) and define
W = {α1 w1 + α2 w2 + α3 w3 ∈ F5 | α1 , α2 , α3 ∈ F}.
We claim U ⊕ W = F5 . There are three things to prove: (1) W is a subspace of
F5 , (2) U + W = F5 , and (3) this sum is direct.
To see that W is a subspace of F5 , note that 0 · w1 + 0 · w2 + 0 · w3 = 0, and
hence 0 ∈ W . Next suppose a, b ∈ W . Then there exist some αk , βk ∈ F, where
k = 1, 2, 3, such that a = α1 w1 + α2 w2 + α3 w3 and b = β1 w1 + β2 w2 + β3 w3 .
But then a + b = (α1 + β1 )w1 + (α2 + β2 )w2 + (α3 + β3 )w3 , which is again
in W , and hence W is closed under addition. Finally, let γ ∈ F. Then γa =
γ(α1 w1 + α2 w2 + α3 w3 ) = (γα1 )w1 + (γα2 )w2 + (γα3 )w3 which is again in W ,
and hence W is closed under scalar multiplication. So W is indeed a subspace.
We next show that U +W = F5 . First notice that U +W ⊆ F5 since U, W are
both subspaces of F5 . To see the that F5 ⊆ U +W , let a = (a1 , a2 , a3 , a4 , a5 ) ∈ F5 .
Recalling our definition of the vectors v1 , v2 , w1 , w2 , w3 , consider the linear
combination
(a1 v1 + a2 v2 ) + (a3 − a1 − a2 )w1 + (a4 − a1 + a2 )w2 + (a5 − 2a1 )w3 .
Note that the sum above is an element of U + W . And after reducing, we find
that the sum above equals (a1 , a2 , a3 , a4 , a5 ), and hence a ∈ U + W and so in
fact F5 = U + W .
Lastly we show that the sum is direct. Every element of U + W has the form
α1 v1 + α2 v2 + α3 v3 + α4 v4 + α5 v5 for some αk ∈ F with k = 1, . . . , 5, so suppose
0 = α1 v1 + α2 v2 + α3 v3 + α4 v4 + α5 v5 . Simplifying yields
(α1 , α2 , α1 + α2 + α3 , α1 − α2 + α4 , 2α1 + α5 ) = 0.
Clearly α1 = α2 = 0. But now this equation simplifies to
(0, 0, α3 , α4 , α5 ) = 0,
and so α3 = α4 = α5 = 0 as well, and hence the sum is indeed direct.
12
Problem 23
Prove or give a counterexample: if U1 , U2 , W are subspaces of V such
that
V = U1 ⊕ W and V = U2 ⊕ W,
then U1 = U2 .
Proof. The statement is false. Let V = R2 , W = R × {0}, U1 = {0} × R, and
U2 = {(x, x) ∈ R2 | x ∈ R}. Then clearly
V = U1 + W = U2 + W.
Moreover, U1 ∩ W = {0} and U2 ∩ W = {0}, and hence the sums are direct.
That is,
V = U1 ⊕ W = U2 ⊕ W,
but U1 6= U2 .
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